433. Laying Off the Load Line. We shall assume that the arch carries a filling of earth or cinders weighing 100 pounds per cubic foot, that the top of this filling is level, and that it has a thickness of one foot above the crown. • Since concrete weighs about 150 pounds per cubic foot, we shall assume this weight of 150 pounds as the unit of measurement, and therefore reduce the ordinates of earthwork to the load line for the top of the earth, as shown in Fig. 232. We shall assume as an additional dead load a pavement weighing SO pounds per square foot, and shall therefore lay off an ordinate of of foot above the ordinates for the earth-filling load. For this particular problem, we shall only investigate a live load of 200 pounds per square foot, extending over one-half of the span from the abutment to the center. From our previous work in arches, we know that such a loading will test the arch more severely than a similar unit live load extending over the entire arch; and therefore, if the arch proves safe for this eccentric load, we may certainly assume that it will be safe for a full load. These load lines are laid off similarly to the method elaborated in Article 409. The arch has already been laid off in equal horizontal sections, each having a width of 3.07 feet. The two end sections are slightly longer if we consider the entire load which is vertically over the extreme ends of the extrados of the arch. The 1S sections lying between the end sections, have a width of 3.07 feet, and a variable height which may be considered as extending from the top of the load line down to the intrados. We may therefore multiply the widths of these sections by their various heights, and by 150, and obtain the number of pounds weight on each section, and we find the loads as follows: The sum total of this loading, which represents the total dead and live load on a section of the arch one foot wide (in the direction of the axis of the arch), is 63,814 pounds. We lay off these various loads on the right-hand side of the drawing in a vertical line, using a scale of 5,000 pounds per inch. Selecting a pole o, at random, we draw rays to the various points in the load line.
434. Trial Equilibrium Polygon. Commencing at the point 0, we draw the segments of the trial equilibrium polygon parallel with the rays in the force diagram which run from the point 0„ to the load line, and obtain the trial equilibrium polygon OB,. By drawing from o, the line 0 parallel to the line OB„ we obtain the point n on the load line, from which we draw an indefinite horizontal line which will be the locus of the pole of the true equilibrium polygon. A vertical from the point intersects the horizontal from n in the point and this would be the pole of a trial equilibrium polygon whose closing line is horizontal, and whose vertical ordinates are equal to those of the corresponding vertical ordinates of the trial equilibrium polygon It is only necessary to find the proper ratio by which these several ordinates should be multiplied, in order to find the corre sponding ordinates of the special equilibrium polygon. It is also Necessary to shift the entire trial equilibrium polygon up or down, so that the line em"' which corresponds to it shall coincide with the line vm which has already been drawn. The special line /,"'In"'
corresponding to this trial equilibrium polygon, is found by satisfying Equation 56; but since dx is in this case a constant, it is found by determining the average value of z'ff, which is the distance from any point in the trial equilibrium polygon to the proper position of the line ent" corresponding to this equilibrium polygon. If the trial equilibrium polygon had been again redrawn with as a pole, it should terminate in the point B; but it is practically unnecessary to do this, since we may draw the line vm' parallel to and measure the distance from vm' down to the various points of the trial equilib rium polygon These various distances in the column headed are as given in the accompanying tabular form (page 414).
We have here the rather unusual case that the trial equilibrium polygon is entirely below the line vm'; and the ordinates are all negative, instead of being partially positive and partially negative. In any case, the algebraic sum should be taken, which should be divided by the number of ordinates.. In this case we find that the mean value of z'n is —3.44. Drawing the line v"in"' parallel to yin', and at a vertical distance of 3.44 feet below it, we find the position of the vm line for the trial equilibrium polygon OB,. This line in tersects the trial equilibrium polygon at the points h" and k". The student should note that it is a mere accidental coincidence that the point k" comes almost exactly on the intrados. By drawing ver ticals from h" and k" to the line vm, we obtain the points h and k, which will be two points in the true equilibrium polygon.
435. Pole Distance of the True Equilibrium Polygon. It is necessary to satisfy Equation 58. We shall consider in this case that the various points in the arch-rib curve and in the special equilibrium polygon are the points where these two lines are intersected by the load lines through the centers of the 20 sections. Therefore the points where these verticals intersect the center of the arch rib, give the points from which we measure down to the line OB, and obtain various values for y which are given in the tabular form above (Article 434). From Fig. 22S we may observe that z' equals the numerical difference between the value of y and the distance from OB up to the line vm. This distance has already been computed at 10.95 feet. Therefore, having scaled off as accurately as practicable the various values of y, it is unnecessary to scale off the values of z', but merely to take the numerical difference (carefully observing the algebraic sign) between 10.59 and the various values of y. We thus obtain the values of z as given in the tabular form. Since z' measures the ordi nates to points in the curve, and since the curve is symmetrical about its center, it is unnecessary, in this case, to set down the values of z' on both sides of the center; and therefore only the values from 1 to 10, inclusive, are written down in the tabular form. Multiplying the corresponding values of y and z', we find the products as given under the heading yz'. Adding these products for the half-span of the arch, we find an algebraic summation of ± 196.06; multiplying this by 2, we find that the algebraic stumnation for the entire arch is + 392.12.